Advanced Operator Math

transcript

ADVANCED MATH for WATER SYSTEM

OPERATORS

Peter T Nathanson, PENew Mexico Rural Water Association3413 Carlisle Blvd NE Albuquerque, New Mexico 87110505.884.1031

References Basic Math Concepts for Water and Wastewater Plant Operators, Joanne Kirkpatrick Price, Technomic Publishing Co., Inc., 1991.

Applied Math for Water/Wastewater Plant Operators & Workbook, Texts and Workbooks, Joanne Kirkpatrick Price, Technomic Publishing Co., Inc., 1991.

Wastewater Math The Basics, Skeet Arasmith, ACR Publications, Inc., 1995.

The Math Text for Water and Wastewater Technology, 2nd ed., Grover Wright, Wright’s Training, 1994.

Simplified Math for Waterworks Operators, George Mason, ACR Publications, Inc., 1992.

Topics Conversions

Detention Time

Flow and Velocity

Chemical Dosage

Percent Strength

Pressure and Head

Pumping System Hydraulics

Pumping System Costs

Abbreviations and AcronymsA = areaV = velocityt = timeVol = volume# = poundsW = widthD = depthL = lengthQ = flowr = radiusdia = diametercir = circumfenceπ = 3.14Hp = horse power

psi = pounds per inchft = feetsq = squarecu = cubicyd = yardDT = detention timemg/L = milligrams per litreppm = parts per millionMGD = million gallons per daygpm = gallons per minuteCl2 = chlorine

Equivalents• 12 inches = 1 foot• 36 inches = 3 feet or 1 yard• 144 square inches = 1 sq. ft.• 9 square feet = 1 sq. yd.• 43,560 sq. ft. = 1 acre• 325,828 gallons = 1 acre foot• 1 cubic foot = 1,728 cu. in.• 1cu.ft. of water contains 7.48 gals. & weighs 62.4 lbs.• 1 gal. of water weighs 8.34 lbs.• 1 liter = 1,000 milliliters• 1 gram = 1,000 milligrams• 1 mg/L = 1 ppm• 1 kilogram = 1,000 grams• 1 pound = 453.6 grams• 1 gal. of water = 3.785 liter or 3,785 milliliters

1 grain per gal. = 17.1 ppm1 cubic yard = 27 cubic feet1% = 10,000 mg/L1 psi = 2.31 feet of water1 atmosphere = 14.7 psi1day = 24 hours = 1,440 min. = 86,400 sec.1 MGD = 694 gals./min = 1,545 ft³/sec.1 Hp = .746 kw = 550 ft. lbs/sec = 33,000ft.lbs./min.

Equivalents (cont’d)

Words and SymbolsHierarchy of Operations

MATH OPERATION SYMBOL EXAMPLE

Multiplication X Q = V x AMultiplication . Q = V . AMultiplication No space Q = VAMultiplication ( ) ( ) Q = (V) (A)Division ÷ r = D ÷ 2Division — r = D

2Division / r = D/2

Word ProblemsWord problems are a series of expressions that fits into an equation and an equation is a combination of math expressions.

Suggestions:• Read the problem entirely to get a feel for the whole problem • Draw a diagram to describe the problem statement• List information and the variables you identify• Attach units of measure to the variables (gallons, miles, inches, etc.) • Define what answer you need, as well as its units of measure • Set up equation(s), solve for variable, populate with data• Work in an organized manner

– Draw and label all graphs and pictures clearly – Note or explain each step of your process; this will help you track variables and remember their meanings

• Look for "key" words that indicate certain mathematical operations.

Common Conversions1. Linear Measurements 1 inch = 2.54 cm 1 foot = 30.5 cm 1 meter = 100 cm = 3.281 =

39.4 inches 1 acre = 43,560 sq. ft. 1 yard = 3 feet2. Volume 1 gal. = 3.78 liters 1 cubic foot (ft³) = 7.48 gals. 1 liter = 1000 mL 1 acre foot = 43,560 cubic feet 1 gal = 16 oz. dry wt.

3. Weight 1 ft³ of water = 62.4 lbs. 1 gal. = 8.34 lbs. 1 lb. = 453.6 grams 1 kg = 1000 g = 2,2 lbs. 1 % = 10,000 mg/L4. Pressure 1 ft. of head = 0.433 psi 1 psi = 2.31 ft. of head5. Flow 1 cfs = 448 gpm 1 gpm = 1440 gpd

Example # 1• Question: How many feet are in 18 inches• Known: 1 foot = 12 inches• Solution?

Example # 21. Question: How many gallons are in

3291 cubic feet (cu.ft.)?2. Known: 1 cubic feet = 7.48 gallons.3. Solution: ?

Example # 31. Question: how many feet are in ¼ mile?2. Known: 1 mile = 5280 ft.3. Solution: ?

Example # 4• Question: convert 3,920 cu.ft. to cubic yards

(cu.yds.)• Known: 1 cu.yds. = 27 cu.ft.• Solution: ?

Example # 5• Gallons Per Day (GPD)• Million Gallons Per Day (MGD)

• Question: convert 3,211,000 GPD to MGD • Known: 1 MGD = 1,000,000 GPD• Solution: ?

The Ohm’s Law Pie Chart

The Ohm’s Law Pie ChartShortcut Calculations

Current, I (Amps)“Flow” of electricity defined as

one Coulomb per second

Voltage, V (Volts)Defined as Electromotive Force, or EMF

Similar to pressure in a water system

Resistance, R (Ohms) The unit of resistance to current flow – similar to

headloss in a water system

An ohm is the amount of resistance that allows 1 amp of current to flow when the applied voltage is 1 volt

Power, P (Watts or HP) A function of both voltage and amps:Volts X Amps = Watts

Wattage is a measure of work

1000 watts = 1 KW = 1.34 HP, or

1 HP = 746 watts = 0.746 KW

Temperature ConversionThere are 2 scales used to report temperature:• Fahrenheit ( F° ) = English scale• Celsius ( C° ) = metric scale• C° = 5/9 ( F°- 32° ) or • C° = 0.55 ( F°- 32° ) or• C° = (F°- 32°) ÷ 1.8• F° = (9/5 x C°) + 32° or• F° = (1.8 x C°) + 32°

DETENTIONTIME

Calculating Detention Time (DT)• Use – detention time is the amount of time that a

fluid stays in a container.• Units – detention time is expressed in units of

time. The most common are seconds, minutes, hours and days.

• Calculations divide the volume of the container by the flow rate.

• Volume units are gallons or sometimes cubic feet.• Time units – will be whatever units are used to

express the flow;– GPM=DT(min), GPD=DT(days)

Detention Time (DT)• DT = volume of tank

flow, gals/time

• A chlorine contact chamber holds 5,000 gals. It is desired to have a contact time (CT) of 30 minutes in the chamber. What is the maximum flow rate that can pass through this chamber at this DT?

• DT = Vol (place the known values in the equation)flow

• 30 min = 5000 gals gal/min

• Rearrange the equation• GPM = 5000 gals or 166.66 gpm

30 min

5000 galsFlow in

Flow out

Example

• A water reservoir that is 20-ft in diameter with a depth of 16-ft needs to be filled. If a well is pumping at 200 gpm, how many minutes will it take to fill the tank? How many hours?

FLOWand

VELOCITY

Flow RatesFlow is expressed in:• Gallons per minute (gpm)• Cubic feet per second (cfs)• Gallons per day (gpd)• Million gallons per day (mgd)

Conversions:• 1 cfs = 448 gpm• 1 gpm = 1440 gpd• mgd = gpd ÷ 1,000,000

Flow RatesFlow in a pipeline, channel or stream at a

particular moment depends on the cross-sectional area and the velocity of water moving through it & is found using the equation:

• Q = A x V • Q = flow rate, vol per time• A = area (L x L)• V = velocity, L/time• If a circular pipe is flowing full (most situations)

the resulting flow rate is expressed as• Q = 0.7854 x D² x V

Example• Find the flow in cfs in a 6 inch pipe if the velocity

is 2 feet per second.

Example• A 15 in. diameter pipe is flowing full. What is the

gallons per minute flow rate in the pipe if the velocity is 110 fps?

Example• A channel is 3 ft. wide has water flowing at a

depth of 2 ft. The velocity in the channel is found to be 1.8 fps. What is the cubic feet per second flow rate in the channel?

1.8 fps

CHEMICALDOSAGE

Chemical Dosage

# or #/day = Vol (MG) or Flow (MGD) x 8.34 x Concentration (mg/L)

Volume always in million gallons (MG, or Mgal)Flow always in million gallons per day (MGD)

Concentration always in milligrams per liter (mg/L) = ppm8.34 = weight of water = 8.34 #/gallon

# = V, MG x C, mg/L x 8.34, or#/day = Q, MGD x C, mg/L x 8.34

Concept: Converts concentration to pounds:Chlorine, fluoride, copper sulfate, orthophosphate,

nitrate, iron, etc….

Milligrams-per-litre to Pounds-per-day

• Continuous feed for water supply#/day = Q, MGD x dose, mg/L x 8.34 lb/gal

• One-time feed for tank, pipe, well# = Vol, Mgal x dose, mg/L x 8.34 lb/gal

Concentration to MassShortcut Calculations

Q 8.34

V 8.34

Chemical DosageA water plant’s daily flow is 10,000 gallons. The chlorine dose is 1.5 mg/L. How may pounds of chlorine is being fed per day?

10,0001,000,000

x 8.34 x 1.5 = 0.1251 lbs chlorine/day

Lbs/day = Flow (MGD) x 8.34 x Concentration (mg/L)

How many gallons of Sodium Hypochlorite is needed? Assume a 5% available chlorine solution.

= 2.502 lbs of NaOCl0.1251.05

2.502 lbs8.34 lbs/gal = 0.3 gals per day

Chemical DosageA new water storage tank needs to be disinfected prior to use. The new tank capacity is 15,000 gallons. How many gallons of 7% sodium hypochlorite will be needed to disinfect the tank with a 0.5 mg/L chlorine residual? Step 1: How many lbs of chorine will be needed?

15,0001,000,000

x 8.34 x 0.5 = 0.06255 lbs chlorine

Step 2: How many lbs of NaOCl will be needed?

= 0.893571 lbs of NaOCl0.062550.07

Step 3: How many gallons of NaOCl will be needed?

= 0.107143 gallons of NaOCl0.8935718.34

Note: 1 pint = 0.125 gallons1 liter = 0.26417 gallons

PERCENT STRENGTH of

SOLUTIONS

Percent Strength

1. The percent strength of a solution can be expressed as percent-by weight.

2. % strength (by Wt.) = Wt. of Solute X 100%Wt. of Solution

Solute = Wt. of Chemical Solution = the combined Wt. of Solute and Solvent

• If 25 lbs. of chemical is added to 400 lbs. of water, what is the percent strength of the solution by weight?

• Wt. of Solution = Wt. of Solute + Wt. of Solvent• = 25 lbs. + 400 lbs. • = 425 lbs.

Percent Strength

• % Strength (by Wt.) = Wt. of Solute X 100%Wt. of Solution

= 25 lbs. X 100%425 lbs.

= 0.059 X 100%

= 5.9 %

Percent Strength• If 40 lbs. chemical is added to 120 gals. of water, what is

the percent strength of the solution by weight?1. % strength (by Wt) = Wt of Solute X 100%

Wt of Solution Solute = Wt. of Chemical Solution = the combined Wt of Solute and Solvent Convert 120 gals. of water to pounds of water. (120 lbs) (8.334lbs/gal) = 1001 lbs

Wt. of Solution = Wt. of Solute + Wt. of Solvent = 40 lbs + 1001lbs = 1041 lbs

Percent Strength

• % Strength (by Wt.) = Wt. of Solute X 100%Wt. of Solution

= 40 lbs X 100%1041 lbs

= 0.038 X 100%

= 3.8 %

PRESSUREand

FEET OF HEAD

PRESSURE VS FEET OF HEAD

Pressure & Head Calculations• Pressure is the weight per unit area• Pounds per square inch, lbs/in²• Pounds per square foot, lbs/ft²• Pressure on the bottom of a container is not

related to the volume of the container, nor the size of the bottom.

• Pressure is dependant on the height of the fluid in the container.

• The height of the fluid in a container is referred to as Head. Head is a direct measurement in feet & directly related to pressure.

Relationship between Feet & Head• Weight of Water is 62.4 pounds per cu.ft.• 7.48 gal/cuft x 8.34 lbs/gal = 62.4 lbs/cuft

1cuft1 ft

1 ft 1 ft.

1cuft1 ft

• Weight of Water is 62.4 pounds per cu.ft.• 7.48 gal/cuft x 8.34 lbs/gal = 62.4 lbs/cuft

1cuft1 ft

1 ft.1 ft

1cuft1 ft

1 ft.1 ft

1 ft 1cuft

1 ft.1 ft

Pressure & Head• Imagine a cube of water 1ft x 1ft x 1ft. Then, the surface

area of any one side of the cube will contain 144 in² (12in x 12in = 144 in²). The cube will also contain 144 columns of water one foot tall & one inch square.

1in1inWeight = 62.4lbs/144in²

= 0.433lbs/in² or= 0.433 psi

Therefore, 1 Feet of head= 1ft 0.433psi

= 2.31 ft/psi

So, 1ft = 0.433 psi, and1psi = 2.31 feet

1in1in

Remember these conversion factors

• PSI x 2.31 ft/psi = FEET

• FEET x 0.433 psi/ft = PSI

• Examples…

PRESSURE GAUGES

• PSIG = PSI GAUGE

• PSIA = PSI ABSOLUTE

HYDRAULICSof

PUMPING SYSTEMS

Static Head• In a system where the reservoir feeding the pump is

higher than the pump, the difference in elevation between the pump centerline and the free water surface of the reservoir feeding the pump is called static suction head

Reservoir1

Reservoir2

Total statichead

Pump CLPump CL

Static discharge head

Reservoir1

Pump CL

Reservoir2

Reservoir1

Pump CL

Static suction head

Reservoir2

Reservoir1

Static head

• Static Discharge Head is defined as the difference in height between the pump centerline and the level of the free water surface on the discharge side of the pump.

• Total Static Head is the total height that the pump must lift the water when moving it from reservoir #1 to reservoir #2.

Static Lift• In a system where the reservoir feeding the pump is

lower than the pump, the difference in elevation between the centerline and the free water surface of the reservoir feeding the pump is called static suction lift

Pump CL

Total static headStatic discharge head

Static suctionlift

Example• Locate, label and calculate (in feet):• Static suction head or static suction lift• Static discharge head• Total static head

Reservoir1

Reservoir2

Pump CL

742 ft

722 ft

927 ft

NET POSITIVE SUCTION HEAD, NPSH• NPSH is the total energy available to move water

into the volute and the eye of the impeller (max lift)• At sea level NPSH = 1 atm = 14.7 psi = 34 ft• However, this energy is reduced by:

– static suction lift (biggest factor)– velocity head, VH to get water moving– head loss, HL

– Vapor pressure (based on temperature)• Portion of water evaporates when placed under

a vacuum (at the eye of impeller)• NPSHR = net positive suction head required• NPSHA = net positive suction head available

CHANGES IN ATMOSPHERIC PRESSURE WITH HEIGHT

Vapor Pressure vs TemperatureTemperature, oF Vapor Pressure, feet

32 0.204

59 0.565

68 0.774

100 2.17

150 8.56

200 26.45 58

Total Dynamic Head, TDH Total energy in feet required to move water from fluid level

suction side to fluid level discharge sidecombination of EH, VH and HL of suction and discharge linesVH discharge side > VH suction side due to reduced diameter of

discharge piping Total dynamic suction lift, TDSL fluid level suction side below eye of impellerTDSL = static suction lift + VH + HLCompare to net positive suction head, NPSH – energy reqd to

move water into volute or impeller eyeavailable NPSH = Patm – Pv – static suction lift – HLCheck pump curve to compare available vs required NPSH

Total dynamic suction head, TDSH fluid level suction side above eye of impellerNPSH = static suction head + Patm – Pv – HLpump choice ok if static suction head ≥ NPSH requiredTDSH = static suction head + Patm – VH – HL

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PUMPING SYSTEMCOSTS

Pump and Motor Efficienciesmotor or wire HP, MHP = electrical energy in HP supplied to motor; motor efficiency determines brake HP

brake HP, BHP = mechanical energy in HP supplied to pump shaft from motor; pump efficiency determines water HP

water HP, WHP = mechanical energy in HP transferred to water by pump

Motor,% efficiency

Pump,% efficiencyMHP BHP WHP

Horsepower Requirements Water HP – energy transferred to water by pump HP = (Q, gpm x 8.34 #/gal x TDH, ft)/33,000 ft-#/minWHP = (Q x TDH)/3960

Brake HP – energy transferred to shaft of pump from shaft of motor Brake HP = WHP/Effpump

Motor or Wire HP – energy required in electrical input to the motorMotor or Wire HP = BHP/Effmotorused to calculate cost of pump operation

Advanced Operator Math

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